Question: The equation of hyperbola $H$ is $\dfrac {(x+6)^{2}}{25}-\dfrac {(y+6)^{2}}{4} = 1$. What are the asymptotes?
We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+6)^{2}}{4} = - 1 + \dfrac {(x+6)^{2}}{25}$ Multiply both sides of the equation by $4$ $(y+6)^{2} = { - 4 + \dfrac{ (x+6)^{2} \cdot 4 }{25}}$ Take the square root of both sides. $\sqrt{(y+6)^{2}} = \pm \sqrt { - 4 + \dfrac{ (x+6)^{2} \cdot 4 }{25}}$ $ y + 6 = \pm \sqrt { - 4 + \dfrac{ (x+6)^{2} \cdot 4 }{25}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 6 \approx \pm \sqrt {\dfrac{ (x+6)^{2} \cdot 4 }{25}}$ $y + 6 \approx \pm \left(\dfrac{2 \cdot (x + 6)}{5}\right)$ Subtract $6$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{2}{5}(x + 6) -6$